18.755 Problem Set 2 solutions
1. Suppose f ∈ C ∞ (R), and f (0) = 0. Prove that there is a function g ∈ C ∞ (R)
with the property that
g(x) = f (x)/x
(x 6= 0).
What is g(0) (in terms of f )?
The hint to consider
G(x) =
1
Z
f ′ (tx)dt
0
was the whole ball game. If x = 0, the integrand is constant, and we get
G(0) = f ′ (0).
If x 6= 0, we can do the change of variable s = tx and compute
Z x
ds
f (x) − f (0)
f (x)
f ′ (s)
G(x) =
=
=
.
x
x
x
0
So G agrees with g for x 6= 0, and G(0) = f ′ (0). To see that G is smooth, you can quote a
general fact: if h(x, t) is smooth on [0, 1] × U (with U an open set in R), then
H(x) =
Z
1
h(x, t)dt
0
is a smooth function of x ∈ U . More precisely,
dn H
=
dxn
Z
0
1
∂ n H(x, t)
dt.
∂xn
That is, you can differentiate under the integral sign. (To say that a function is smooth on
[0, 1] means that it is smooth on some open set containing [0, 1]; or, equivalently, that all
the one-sided derivatives at 0 and 1 exist.) Probably you can find this general fact in blue
Rudin; the proof requires showing that
H(x + ǫ, t) − H(x, t)
ǫ
converges to
∂H
∂x (x, t)
uniformly in t ∈ [0, 1] as ǫ tends to zero.
2. Write
m0 = {f ∈ C ∞ (R) | f (0) = 0}.
Prove that
m20 = {h ∈ C ∞ (R) | h(0) = h′ (0) = 0}.
1
2
That any function h ∈ m20 satisfies h(0) = h′ (0) = 0 is an immediate consequence of the
Leibnitz formula for differentiating products. What is not so obvious is the converse. So
suppose that h is smooth, and that
h(0) = h′ (0) = 0.
According to Problem 1, the function
g(x) =
h(x)/x
x 6= 0
h′ (0) = 0 x = 0
is smooth. So the two functions g and x both belong to m; so h = gx belongs to m2 .
(You might have tried to proceed by showing that h has a smooth square root. Of course
there are problems if h takes negative values, but do you think the following statement is
true? If h is a nonnegative smooth function on R, then the nonnegative square root h1/2 is
also smooth.)
3. Let M be the real line; I’ll write x for a variable point of M . Consider the
three vector fields
Xx(0) =
d
,
dx
Xx(1) = x
d
,
dx
Xx(2) = x2
d
.
dx
For every real number t0 and every x0 ∈ M (also a real number) calculate the
(p)
maximal integral curves γx0 ,t0 for each of the three vector fields X (p) (p = 0, 1, 2).
For X (0) , the differential equation is
dγ
= 1,
dt
γ(t0 ) = x0 .
The first condition implies that γ(t) = t + C, and the initial condition gives
(0)
γx0 ,t0 (t) = x0 + (t − t0 )
(−∞ < t < ∞).
For X (1) , the differential equation is
dγ
= γ(t),
dt
γ(t0 ) = x0 .
This is a familiar differential equation from any course on the subject; the first condition
implies that γ(t) = Cet , and the initial condition gives
(1)
γx0 ,t0 (t) = x0 e(t−t0 )
(−∞ < t < ∞).
3
For X (2) , the differential equation is
dγ
= γ 2 (t),
dt
γ(t0 ) = x0 .
If x0 = 0, one obvious solution (and therefore the only one, by the uniqueness theorem
written with the problems) is
γ(t) = 0
(−∞ < t < ∞).
If x0 6= 0, then we can divide by γ 2 (t) at least for t close to t0 and get
dγ
= dt.
γ2
Integrating gives
−1
= t + C,
γ(t)
or equivalently
γ(t) =
−1
.
t+C
Plugging in the initial condition γ(t0 ) = x0 6= 0 gives C = −t0 − x−1
0 , and
(2)
γx0 ,t0 (t)
x0
=
1 − x0 (t − t0 )
−∞ < t < t0 + 1/x0 x0 > 0
t0 + 1/x0 < t < +∞ x0 < 0.
It’s also clear that these paths γ have no continuous extension beyond the indicated range of
(2)
values of t (because γx0 ,t0 (t) tends to ∞ as t approaches t0 + 1/x0 ); so they are the maximal
integral curves.
4. Let M = R2 be the real plane; I’ll write (x1 , x2 ) for a variable point of M .
Consider the three vector fields
= x1
Xx(0)
1 ,x2
∂
∂
+ x2
,
∂x1
∂x2
= x1
Xx(1)
1 ,x2
∂
∂
− x2
,
∂x1
∂x2
= x1
Xx(2)
1 ,x2
∂
∂
− x2
.
∂x2
∂x1
For every real number t0 and every (x1,0 , x2,0 ) ∈ M calculate the maximal integral
curves
(p)
(p)
(p)
γ(x1,0 ,x2,0 ),t0 = (γ1,(x1,0 ,x2,0 ),t0 , γ2,(x1,0 ,x2,0 ),t0 )
for each of the three vector fields X (p) .
If V is an n-dimensional real vector space, we can regard V as an n-dimensional manifold,
and naturally identify each tangent space Tv V with v. A vector field X on V is therefore
4
the same thing as a smooth map from V to V . The three vector fields in the problem come
from three linear maps from V = R2 to itself:
= A(p)
Xx(p)
1 ,x2
A(0) = I2 ,
A(1) =
1
0
0
−1
x1
x2
,
A(2) =
0 −1
1 0
.
The corresponding differential equations
γ ′ (t) = A(p) γ(t)
are therefore linear differential equations (with constant coefficients). The solutions therefore exist for all time, and are given by matrix exponentials:
(p)
γ(x1,0 ,x2,0 ),t0 (t)
(p))
= exp((t − t0 )A
x1,0
x2,0
.
The matrix exponentials can be calculated by finding eigenvectors or just computing the
powers by hand; in any case they are
exp(tA(0) ) = et I2 ,
exp(tA(1) ) =
et
0
0
e−t
,
exp(tA(2) ) =
cos(t)
sin(t)
− sin(t)
cos(t)
.
Plugging these formulas into the general formula above gives
(0)
γ(x1,0 ,x2,0 ),t0 (t) = e(t−t0 ) (x1,0 , x2,0 ),
(1)
γ(x1,0 ,x2,0 ),t0 (t) = (e(t−t0 ) x1,0 , e−(t−t0 ) x2,0 ),
(2)
γ(x1,0 ,x2,0 ),t0 (t) = (x1,0 cos(t − t0 ) − x2,0 sin(t − t0 ), x1,0 sin(t − t0 ) + x2,0 sin(t − t0 )).
Each integral curve is therefore either
(0) the point (0, 0), or a ray emanating from zero; or
(1) the point (0, 0), or half of a coordinate axis, or one branch of the hyperbola xy = A; or
(2) the point (0, 0), or a circle centered at the origin.